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Find the inverse f(x) = x2. Then find the min and max given that -1<=x<1.

User Phininity
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1 Answer

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Final answer:

To find the inverse of f(x) = x^2 within the interval -1<=x<1, we solve for x, which gives f^-1(x) = sqrt(x). The min is f(0) = 0, and the max is f(-1) = 1 in the given interval.

Step-by-step explanation:

To find the inverse of the function f(x) = x2, we need to solve for x when f(x) is given.

Assuming the principal branch of square root for the inverse, the inverse function can be written as f-1(x) = √(x) or x = √(f(x)).

Since we are considering the domain of f(x) to be -1≤x<1, we focus only on the non-negative square root, thus f-1(x) = √(x).

For the given interval, we analyze the original function f(x) = x2.

The minimum value of f(x) occurs at x = 0, which is f(0) = 02 = 0.

Since the function is a parabola opening upwards and the x-values are restricted between -1 and 1, the maximum value of f(x) will occur at the endpoints of the interval,

which is f(-1) = (-1)2

= 1

and f(1) = (1)2

= 1.

However, since x<1, we exclude the value at x=1, and thus, the maximum in the interval is f(-1) = 1.

User Galcyurio
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